Mathematics, 21.10.2019 22:00, isanaty7951
Solve for the roots in the equation below. in your final answer, include each of the necessary steps and calculations. hint: use your knowledge of factoring polynomials and identities for imaginary numbers.
x^3- 27i = 0
Answers: 1
![\bf (x+3i)[x^2-3ix+(3i)^2]=0\implies \begin{cases} x+3i=0\implies \boxed{x=-3i}\\\\ x^2-3ix+(3i)^2=0 \end{cases} \\\\\\ \textit{now, for the second one, we'd need the quadratic formula} \\\\\\ x^2-3ix+(3i)^2=0\implies x^2-3ix+(3^2i^2)=0 ](/tpl/images/0339/3704/9a00c.png)
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Solve for the roots in the equation below. in your final answer, include each of the necessary steps...
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